Problem: Graph this system of equations and solve. $8x-2y = -8$ $2x-2y = 4$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ Click and drag the points to move the lines.
Convert the first equation, $8x-2y = -8$ , to slope-intercept form. $y = 4 x + 4$ The y-intercept for the first equation is $4$ , so the first line must pass through the point $(0, 4)$ The slope for the first equation is $4$ . Remember that the slope tells you rise over run. So in this case for every $4$ positions you move up $1$ position to the right. $4$ positions up from $(0, 4)$ is $(1, 8)$ Graph the blue line so it passes through $(0, 4)$ and $(1, 8)$ Convert the second equation, $2x-2y = 4$ , to slope-intercept form. $y = x - 2$ The y-intercept for the second equation is $-2$ , so the second line must pass through the point $(0, -2)$ The slope for the second equation is $1$ . Remember that the slope tells you rise over run. So in this case for every $1$ position you move up You must also move $1$ position to the right. $1$ position to the right. $1$ position up from $(0, -2)$ is $(1, -1)$ Graph the green line so it passes through $(0, -2)$ and $(1, -1)$ The solution is the point where the two lines intersect. The lines intersect at $(-2, -4)$.